A wave is shown in the diagram. What is the frequency of the wave if its speed is 150 meters per second?
We see in this diagram a picture of the wave. And the displacement of the wave in units of meters is plotted against the distance the wave has travelled horizontally, again in units of meters. Based on the information in this picture as well as what we’re told in the problem statement, we want to solve for the frequency of this wave.
To do this, we’ll relate the wave frequency to the wave speed and wavelength. There is a mathematical equation, we can recall, that help us do that. That equation says that wave speed 𝑣 is equal to wave frequency 𝑓 multiplied by wavelength 𝜆. Now in our case, it’s not the wave speed we want to solve for, but rather the wave frequency. If we divide both sides of our wave speed equation by the wavelength 𝜆, then we find that wave frequency is equal to wave speed divided by wavelength.
In the problem statement, we’re told what 𝑣 is. It’s 150 meters per second. That’s how fast this wave is moving, but what we don’t yet know is the wavelength, 𝜆. However, we can use the information in our graph to find it. First off, let’s recall what a wavelength is in general. A wavelength is the distance along the wave’s direction of motion that’s needed to complete one full cycle of motion.
Considering this wave in our chart then, we could start at the trough of the wave, the lowest point, and one complete cycle would be to the next trough over. Or equivalently, we could start out at a zero point here, and then one wavelength would be the distance from there to another zero point with a similar slope. Or yet, another way to calculate wavelength is to start at a crest, a high point, and then to calculate the distance to the next crest over. All of these distances are the same and they’re all equal to one wavelength of this wave.
So, let’s use one of these measurements to calculate the wavelength of this wave in particular. Let’s see we choose to calculate this wave’s wavelength using this distance here, the distance between these two zero points along the wave. We can see that we start out at a distance of two meters and end up at a distance of six meters. Therefore, the wavelength 𝜆 is equal to six meters minus two meters, or four meters.
And just as a quick check, notice that we would’ve gotten the same result if we had used, say, the distance from one peak to another. In that case, we would calculate the distance between one meter and five meters, which, again, is four meters. So then, our calculation of wavelength seems to be good. Four meters is the wavelength of this particular wave. Knowing that, we can now substitute in four meters for 𝜆 in our equation of frequency. And we can replace the wave speed 𝑣 with the given speed of 150 meters per second.
Now before we calculate this fraction, notice what happens to the units in the expression. In both numerator and denominator, we have units of meters. Therefore, this unit cancels out. And we’ll end up with a unit of inverse seconds, or one over seconds. That unit, inverse seconds, is equivalent to the unit of hertz, symbolized Hz. This means that the answer to our question will be given in these units of hertz.
And if we divide 150 by four, we find a result of 37.5 hertz. Based on the wave speed and wavelength then, every second of time that passes, this wave goes through 37 and a half complete cycles. In other words, moving from crest to crest or from trough to trough or anywhere in between, as long as it goes through one complete cycle. This is the frequency of the wave.